Matrices: knowing that $M^3=N^3$ and that $MN^2=NM^2$, how to prove $M^2+N^2$ is non-invertible?

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The exercise:
Let $M$ and $N$ be two distinct $ntimes n$ matrices such that $M^3=N^3$ and that $MN^2=NM^2$. Prove that $M^2+N^2$ is a non-invertible matrix.

I guess I need to show that the determinant is zero, or create a proof by contradiction, but I can't find a good demonstration.

linear-algebra matrices

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edited Jul 18 at 21:22

amWhy

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asked Jul 16 at 14:00

Alexandre Tourinho

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up vote
3
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The exercise:
Let $M$ and $N$ be two distinct $ntimes n$ matrices such that $M^3=N^3$ and that $MN^2=NM^2$. Prove that $M^2+N^2$ is a non-invertible matrix.

I guess I need to show that the determinant is zero, or create a proof by contradiction, but I can't find a good demonstration.

linear-algebra matrices

share|cite|improve this question

edited Jul 18 at 21:22

amWhy

189k25219431

asked Jul 16 at 14:00

Alexandre Tourinho

1305

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

The exercise:
Let $M$ and $N$ be two distinct $ntimes n$ matrices such that $M^3=N^3$ and that $MN^2=NM^2$. Prove that $M^2+N^2$ is a non-invertible matrix.

I guess I need to show that the determinant is zero, or create a proof by contradiction, but I can't find a good demonstration.

linear-algebra matrices

share|cite|improve this question

edited Jul 18 at 21:22

amWhy

189k25219431

asked Jul 16 at 14:00

Alexandre Tourinho

1305

The exercise:
Let $M$ and $N$ be two distinct $ntimes n$ matrices such that $M^3=N^3$ and that $MN^2=NM^2$. Prove that $M^2+N^2$ is a non-invertible matrix.

I guess I need to show that the determinant is zero, or create a proof by contradiction, but I can't find a good demonstration.

linear-algebra matrices

share|cite|improve this question

edited Jul 18 at 21:22

amWhy

189k25219431

asked Jul 16 at 14:00

Alexandre Tourinho

1305

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 21:22

amWhy

189k25219431

edited Jul 18 at 21:22

amWhy

189k25219431

edited Jul 18 at 21:22

amWhy

189k25219431

189k25219431

asked Jul 16 at 14:00

Alexandre Tourinho

1305

asked Jul 16 at 14:00

Alexandre Tourinho

1305

asked Jul 16 at 14:00

Alexandre Tourinho

1305

1305

add a comment | 

add a comment | 

3 Answers
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up vote
9
down vote

By contradiction, if it exists $A$ such that $(M^2+N^2)A=I$.

Then

$$M=MI=M(M^2+N^2)A=(M^3+MN^2)A=(N^3+NM^2)A=N(N^2+M^2)A=N$$

In contradiction with the hypothesis $M neq N$.

share|cite|improve this answer

edited Jul 16 at 14:32

answered Jul 16 at 14:06

mathcounterexamples.net

24.2k21653

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up vote
3
down vote

Consider $(M-N)(M^2+N^2)=M^3-NM^2+MN^2-N^3$.

Can you finish?

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answered Jul 16 at 14:15

egreg

164k1180187

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  On this way, I think $(M-N)(M^2+N^2)=0$ implies that if $(M^2+N^2)$ were invertible, $(M-N)$ would be $0$ and then $M=N$, a contradiction. Is that right?
  – Alexandre Tourinho
  Jul 16 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AlexandreTourinho Good!
  – egreg
  Jul 16 at 14:27
  

  
  
  
  

add a comment | 

up vote
2
down vote

Starting from the identity $MN^2 = NM^2$:

$$
beginalign*
&Leftrightarrow
MN^2 +M^3= NM^2 +N^3 \ &Leftrightarrow
M(N^2 +M^2)= N(M^2 +N^2) \
endalign*
$$
Now, if $N^2+M^2$ were invertible, we'd arrive at $M=N$.

share|cite|improve this answer

answered Jul 16 at 14:21

Sudix

7911316

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
9
down vote

By contradiction, if it exists $A$ such that $(M^2+N^2)A=I$.

Then

$$M=MI=M(M^2+N^2)A=(M^3+MN^2)A=(N^3+NM^2)A=N(N^2+M^2)A=N$$

In contradiction with the hypothesis $M neq N$.

share|cite|improve this answer

edited Jul 16 at 14:32

answered Jul 16 at 14:06

mathcounterexamples.net

24.2k21653

add a comment | 

up vote
9
down vote

By contradiction, if it exists $A$ such that $(M^2+N^2)A=I$.

Then

$$M=MI=M(M^2+N^2)A=(M^3+MN^2)A=(N^3+NM^2)A=N(N^2+M^2)A=N$$

In contradiction with the hypothesis $M neq N$.

share|cite|improve this answer

edited Jul 16 at 14:32

answered Jul 16 at 14:06

mathcounterexamples.net

24.2k21653

add a comment | 

up vote
9
down vote

up vote
9
down vote

By contradiction, if it exists $A$ such that $(M^2+N^2)A=I$.

Then

$$M=MI=M(M^2+N^2)A=(M^3+MN^2)A=(N^3+NM^2)A=N(N^2+M^2)A=N$$

In contradiction with the hypothesis $M neq N$.

share|cite|improve this answer

edited Jul 16 at 14:32

answered Jul 16 at 14:06

mathcounterexamples.net

24.2k21653

By contradiction, if it exists $A$ such that $(M^2+N^2)A=I$.

Then

$$M=MI=M(M^2+N^2)A=(M^3+MN^2)A=(N^3+NM^2)A=N(N^2+M^2)A=N$$

In contradiction with the hypothesis $M neq N$.

share|cite|improve this answer

edited Jul 16 at 14:32

answered Jul 16 at 14:06

mathcounterexamples.net

24.2k21653

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 14:32

edited Jul 16 at 14:32

edited Jul 16 at 14:32

answered Jul 16 at 14:06

mathcounterexamples.net

24.2k21653

answered Jul 16 at 14:06

mathcounterexamples.net

24.2k21653

answered Jul 16 at 14:06

mathcounterexamples.net

24.2k21653

24.2k21653

add a comment | 

add a comment | 

up vote
3
down vote

Consider $(M-N)(M^2+N^2)=M^3-NM^2+MN^2-N^3$.

Can you finish?

share|cite|improve this answer

answered Jul 16 at 14:15

egreg

164k1180187

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  On this way, I think $(M-N)(M^2+N^2)=0$ implies that if $(M^2+N^2)$ were invertible, $(M-N)$ would be $0$ and then $M=N$, a contradiction. Is that right?
  – Alexandre Tourinho
  Jul 16 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AlexandreTourinho Good!
  – egreg
  Jul 16 at 14:27
  

  
  
  
  

add a comment | 

up vote
3
down vote

Consider $(M-N)(M^2+N^2)=M^3-NM^2+MN^2-N^3$.

Can you finish?

share|cite|improve this answer

answered Jul 16 at 14:15

egreg

164k1180187

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  On this way, I think $(M-N)(M^2+N^2)=0$ implies that if $(M^2+N^2)$ were invertible, $(M-N)$ would be $0$ and then $M=N$, a contradiction. Is that right?
  – Alexandre Tourinho
  Jul 16 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AlexandreTourinho Good!
  – egreg
  Jul 16 at 14:27
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
down vote

Consider $(M-N)(M^2+N^2)=M^3-NM^2+MN^2-N^3$.

Can you finish?

share|cite|improve this answer

answered Jul 16 at 14:15

egreg

164k1180187

Consider $(M-N)(M^2+N^2)=M^3-NM^2+MN^2-N^3$.

Can you finish?

share|cite|improve this answer

answered Jul 16 at 14:15

egreg

164k1180187

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 14:15

egreg

164k1180187

answered Jul 16 at 14:15

egreg

164k1180187

answered Jul 16 at 14:15

egreg

164k1180187

164k1180187

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  On this way, I think $(M-N)(M^2+N^2)=0$ implies that if $(M^2+N^2)$ were invertible, $(M-N)$ would be $0$ and then $M=N$, a contradiction. Is that right?
  – Alexandre Tourinho
  Jul 16 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AlexandreTourinho Good!
  – egreg
  Jul 16 at 14:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  On this way, I think $(M-N)(M^2+N^2)=0$ implies that if $(M^2+N^2)$ were invertible, $(M-N)$ would be $0$ and then $M=N$, a contradiction. Is that right?
  – Alexandre Tourinho
  Jul 16 at 14:27
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @AlexandreTourinho Good!
  – egreg
  Jul 16 at 14:27
  

  
  
  
  

1

1

On this way, I think $(M-N)(M^2+N^2)=0$ implies that if $(M^2+N^2)$ were invertible, $(M-N)$ would be $0$ and then $M=N$, a contradiction. Is that right?
– Alexandre Tourinho
Jul 16 at 14:27

On this way, I think $(M-N)(M^2+N^2)=0$ implies that if $(M^2+N^2)$ were invertible, $(M-N)$ would be $0$ and then $M=N$, a contradiction. Is that right?
– Alexandre Tourinho
Jul 16 at 14:27

@AlexandreTourinho Good!
– egreg
Jul 16 at 14:27

@AlexandreTourinho Good!
– egreg
Jul 16 at 14:27

add a comment | 

up vote
2
down vote

Starting from the identity $MN^2 = NM^2$:

$$
beginalign*
&Leftrightarrow
MN^2 +M^3= NM^2 +N^3 \ &Leftrightarrow
M(N^2 +M^2)= N(M^2 +N^2) \
endalign*
$$
Now, if $N^2+M^2$ were invertible, we'd arrive at $M=N$.

share|cite|improve this answer

answered Jul 16 at 14:21

Sudix

7911316

add a comment | 

up vote
2
down vote

Starting from the identity $MN^2 = NM^2$:

$$
beginalign*
&Leftrightarrow
MN^2 +M^3= NM^2 +N^3 \ &Leftrightarrow
M(N^2 +M^2)= N(M^2 +N^2) \
endalign*
$$
Now, if $N^2+M^2$ were invertible, we'd arrive at $M=N$.

share|cite|improve this answer

answered Jul 16 at 14:21

Sudix

7911316

add a comment | 

up vote
2
down vote

up vote
2
down vote

Starting from the identity $MN^2 = NM^2$:

$$
beginalign*
&Leftrightarrow
MN^2 +M^3= NM^2 +N^3 \ &Leftrightarrow
M(N^2 +M^2)= N(M^2 +N^2) \
endalign*
$$
Now, if $N^2+M^2$ were invertible, we'd arrive at $M=N$.

share|cite|improve this answer

answered Jul 16 at 14:21

Sudix

7911316

Starting from the identity $MN^2 = NM^2$:

$$
beginalign*
&Leftrightarrow
MN^2 +M^3= NM^2 +N^3 \ &Leftrightarrow
M(N^2 +M^2)= N(M^2 +N^2) \
endalign*
$$
Now, if $N^2+M^2$ were invertible, we'd arrive at $M=N$.

share|cite|improve this answer

answered Jul 16 at 14:21

Sudix

7911316

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 14:21

Sudix

7911316

answered Jul 16 at 14:21

Sudix

7911316

answered Jul 16 at 14:21

Sudix

7911316

7911316

add a comment | 

add a comment | 

 

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